Question

What is (i) the molar mass of `"lead nitrate"`, and (ii), the molar quantity of a `42.65*g` mass of this material?

Answer 1

We take the quotient, `"Mass of stuff"/"Molar mass of stuff"`.

Explanation:

The molar mass of `"lead nitrate"` is the sum of the individual atomic molar masses....

And so we gots...`(a)` `{207.2+2xx14.01+6xx15.999}*g*mol^-1=??*g*mol^-1`

From where did I get these atomic masses? Did I know them off the top of my head? From where will YOU get these masses if asked the question in an exam?

And so with respect to `"lead nitrate"` we take the quotient....

`(b)` `n_"lead nitrate"=(42.65*g)/(331.20*g*mol^-1)=0.1288*mol`

And how many moles of oxygen atoms are in this sample?

And please note the dimensional consistency of the answer; we wanted an answer with dimensions of moles. Did we get one?

i.e. `(g)/(g*mol^-1)=cancel(g)/(cancelg*mol^-1)=1/(1/(mol))=mol` as required. Capisce?

Answer 2

(a) & (b) done already, I will do
c) Formula units `=("moles of substance" * N_A)f.u.`

`=color(violet)( 0.129 "moles")*6.02*10^23`
`=color(orange)(7.75*10^22 f.u.`